英文:
panic: runtime error: index out of range [-1]
问题
当我尝试解决LeetCode上的118. 杨辉三角问题时，出现了奇怪的错误。下面的代码可以通过OJ测试：
func generate(numRows int) [][]int {
res := [][]int{}
for i := 0; i < numRows; i++ {
row := []int{}
for j := 0; j < i+1; j++ {
if j == 0 || j == i {
row = append(row, 1)
} else if i > 1 {
row = append(row, res[i-1][j-1] + res[i-1][j])
}
}
res = append(res, row)
}
return res
}
但是我写的代码如下，却出现了panic错误，但基本逻辑是相同的：
func generate(numRows int) [][]int {
res := [][]int{}
for i := 0; i < numRows; i++ {
row := []int{}
for j := 0; j < i+1; j++ {
if j == 0 || j == i {
row = append(row, 1)
}
if i > 1 {
row = append(row, res[i-1][j-1] + res[i-1][j])
}
}
res = append(res, row)
}
return res
}
我使用了if else if结构，它运行良好，但是我使用了两个if条件判断错误。
实际上，它们的逻辑是相同的，但为什么会出错呢？如果您能解决这个问题，我将不胜感激。祝您好运！
英文:
when I try to solve the leetcode problem of 118. Pascal's Triangle https://leetcode.com/problems/pascals-triangle/
it occured strange bug. Below the code can pass OJ.
func generate(numRows int) [][]int {
res := [][]int{}
for i := 0; i < numRows; i++ {
row := []int{}
for j := 0; j < i+1; j++ {
if j == 0 || j == i {
row = append(row, 1)
} else if i > 1 {
row = append(row, res[i-1][j-1] + res[i-1][j])
}
}
res = append(res, row)
}
return res
}
But the code I write like this and it occur panic, but the basic logic is same.
func generate(numRows int) [][]int {
res := [][]int{}
for i := 0; i < numRows; i++ {
row := []int{}
for j := 0; j < i+1; j++ {
if j == 0 || j == i {
row = append(row, 1)
}
if i > 1 {
row = append(row, res[i-1][j-1] + res[i-1][j])
}
}
res = append(res, row)
}
return res
}
I used if else if structure, it works fine, but I use 2 if condition judgment error.
In fact, their logic is the same, but why the error? I would be grateful if you could solve this problem. Good lcuk for you!
答案1
得分: 2
问题是在第二个版本中你使用了两个if条件，而在第一个版本中你使用了一个if else。
但基本逻辑是相同的
不，逻辑是不同的。结果是当j为0时，你尝试执行j-1。
如果你有两个像这样的if条件，程序将分别进入两个块，只要它们各自的条件满足。
if j == 0 || j == i {
row = append(row, 1)
}
// 只要 i > 1，即使 j 是 0，你仍然会进入这个块
if i > 1 {
row = append(row, res[i-1][j-1] + res[i-1][j])
}
如果第一个if条件满足，你可以使用continue跳过这部分。
if j == 0 || j == i {
row = append(row, 1)
// 继续下一次迭代
continue
}
if i > 1 {
row = append(row, res[i-1][j-1] + res[i-1][j])
}
也就是说，在你的代码的第一个版本中使用if else似乎是合理的。我不确定为什么你想要改变它。
英文:
The problem is that you use 2 if conditions in the second version, while in the first version you have an if else.
> but the basic logic is same
No, the logic is not the same. The result is that you try to do j-1 when j is 0.
If you have 2 if conditions like this, the program will enter both blocks individually if their respective condition is met.
if j == 0 || j == i {
row = append(row, 1)
}
// if j is 0 you still enter this block as long as i > 1
if i > 1 {
row = append(row, res[i-1][j-1] + res[i-1][j])
}
You could use continue to skip this section, if the first if is met.
if j == 0 || j == i {
row = append(row, 1)
// continue with the next iteration
continue
}
if i > 1 {
row = append(row, res[i-1][j-1] + res[i-1][j])
}
That said, using the if else in the first version of your code seems reasonable. I am not sure why you would want to change it.
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